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I wonder if there is some asymptotics for such sum: $ \sum_{p=2}^{n} \frac{1}{p}$, where the sum is taken over all primes of form $ 4k+3 $?

It's well-known that $ \sum_{p=2}^{n} \frac{1}{p}$, where the sum is taken over all primes is diverges and asymptotically is like $ \ln\ln n $.

But I don't even know how to prove that the first sum is diverges.

I am also interested in asymptotics of the number of primes of the form $ 4k+3 $ less than $n$.

Thank you very much!

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I think you mean it is asymptotic to $\log\log n$. –  Pedro Tamaroff Dec 22 '12 at 20:41
    
Thank you, sure I did. =) –  Sergey Finsky Dec 22 '12 at 20:50

1 Answer 1

up vote 5 down vote accepted

A more general version of this follows from Dirichlet's theorem on arithmetic progressions. In general, $$\sum_{p \equiv a \pmod d} \dfrac1p \sim \dfrac1{\phi(d)}\sum_{p} \dfrac1p$$ where $\phi$ is the Euler's totient function. More details can be found here, here, here and here.

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It's nice to have proof the thing that ought to be true really is true! –  Hurkyl Dec 22 '12 at 20:57
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Thank you very much for your answer, but I have one question left: in papers it's proven that $ \sum_{p \equiv a \pmod d} \frac{1}{p^{s}} \sim \dfrac1{\phi(d)}\sum_{p} \frac{1}{p^{s}} $ when $ s \to +1 $. But how does original problem follows from this? (It is not obvious) –  Sergey Finsky Dec 23 '12 at 7:06
    
@SergeyFinsky We know that $\sum_{p \leq n}\dfrac1p \sim \log(\log n)$. Hence, $\sum_{\left(p \equiv a \pmod d \right) \leq n} \sim \dfrac{\log(\log n)}{\phi(d)}$ –  user17762 Dec 23 '12 at 7:08
    
Yes, but in the paper we consider $L$-function, which is but it's definition a series (not finite), so we have to somehow jump from the infinite series to finite. (It's easy to do when the convergence in series is uniform, but it's obviously not the case) I mean, your previous comment is obviously true, but why $ \sum_{p \equiv a \mod d \le n} \frac{1}{p} \sim \frac{\log(\log(n))}{\phi (d)} $. Do you see what I mean? –  Sergey Finsky Dec 23 '12 at 7:16
    
Sorry for not responding for such a long time, I kept forgetting to respond that your answer solves original question, which can be easily deduced from standart $\delta - \epsilon$ arguments. –  Sergey Finsky Feb 11 '13 at 13:53

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